Question:
A student plots two different graphs using the same photoelectric experiment data.
1. Maximum kinetic energy (\(K_{max}\)) against frequency (\(\nu\))
2. Stopping potential (\(V_{o}\)) against frequency (\(\nu\))
Let the slope of graph 1 be \(S_{1}\) and the slope of graph 2 be \(S_{2}\).
What is the relationship between the slopes?
(h = Planck's constant, e = charge on electron)
A student plots two different graphs using the same photoelectric experiment data.
1. Maximum kinetic energy (\(K_{max}\)) against frequency (\(\nu\))
2. Stopping potential (\(V_{o}\)) against frequency (\(\nu\))
Let the slope of graph 1 be \(S_{1}\) and the slope of graph 2 be \(S_{2}\).
What is the relationship between the slopes?
(h = Planck's constant, e = charge on electron)
1. Maximum kinetic energy (\(K_{max}\)) against frequency (\(\nu\))
2. Stopping potential (\(V_{o}\)) against frequency (\(\nu\))
Let the slope of graph 1 be \(S_{1}\) and the slope of graph 2 be \(S_{2}\).
What is the relationship between the slopes?
(h = Planck's constant, e = charge on electron)
Show Hint
Write Einstein's equation, then use the relation between maximum kinetic energy and stopping potential to find both slopes.
Updated On: Nov 12, 2026
- \(\frac{S_{1}}{S_{2}}=h\)
- \(S_{1}\cdot S_{2}=h\)
- \(\frac{S_{1}}{S_{2}}=e\)
- \(S_{1}+S_{2}=e\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Einstein's photoelectric equation is \(K_{max} = h\nu - \phi\).
The stopping potential is related to it by \(K_{max} = eV_{o}\).
Step 2: Slope of graph 1
\(K_{max}\) versus \(\nu\) is a straight line with slope \(S_{1} = h\).
Step 3: Slope of graph 2
\(V_{o} = \dfrac{h}{e}\nu - \dfrac{\phi}{e}\), so the slope of \(V_{o}\) versus \(\nu\) is \(S_{2} = \dfrac{h}{e}\).
Step 4: Relation
\(\dfrac{S_{1}}{S_{2}} = \dfrac{h}{h/e} = e\).
Final Answer:
The ratio of the two slopes equals the electron charge e. \[ \boxed{\text{(C) }\dfrac{S_{1}}{S_{2}}=e} \]
Einstein's photoelectric equation is \(K_{max} = h\nu - \phi\).
The stopping potential is related to it by \(K_{max} = eV_{o}\).
Step 2: Slope of graph 1
\(K_{max}\) versus \(\nu\) is a straight line with slope \(S_{1} = h\).
Step 3: Slope of graph 2
\(V_{o} = \dfrac{h}{e}\nu - \dfrac{\phi}{e}\), so the slope of \(V_{o}\) versus \(\nu\) is \(S_{2} = \dfrac{h}{e}\).
Step 4: Relation
\(\dfrac{S_{1}}{S_{2}} = \dfrac{h}{h/e} = e\).
Final Answer:
The ratio of the two slopes equals the electron charge e. \[ \boxed{\text{(C) }\dfrac{S_{1}}{S_{2}}=e} \]
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